# At a restaurant, you can choose from 2 appetizers, 4 main courses, and 3 desserts. How many different meals, each consisting of one appetizer, one main course, and one dessert, can be created?

##### 2 Answers

#### Answer

#### Answer:

#### Explanation

#### Explanation:

#### Answer:

24

#### Explanation:

First you need to understand the process of selection and arrangement.

Suppose you have 4 numbers - 1, 2, 3, 4 and you have to form 3 digit numbers with all these numbers atmost once (For example 111 is not permitted). How would you do so?

So first step, we have to SELECT 3 numbers from this pile.

We can either choose 1,2,3 or 1,2,4 or 2,3,4 and so on. But how do you definitively know the selections. Here comes in the concept of combinations.

The basic guiding formula for combinations is that

If you have

where n! means product of all natural numbers from n to 1. (for eg. if n = 5 then n! =

Here n = 4 and r = 3 so

Now, you need to ARRANGE these numbers. For this concept, permutations come into play, but I'll leave that for another time as it is not relevant to the question asked by you.

In your question, I just have to SELECT

- 1 appetizer from 2 appetizers

AND - 1 main course from 4 main courses

AND - 1 dessert from 3 desserts

Notice how I have capitalized AND. It is because AND is a key word here.

Now what we have to do is get individual selection from the 3 cases and multiply them.

Remember one thing , AND means multiply and OR means add (For eg. if you had said to choose either one main course OR one dessert, then we would have added the selections)

For 1. , selections =

For 2. , selections =

For 3. , selections =

Total possibilites =

#### Answer

#### Answer:

#### Explanation

#### Explanation:

Describe your changes (optional) 200

We have to choose 1 appetizer out of 2 appetizers =

We have to choose 1 main course out of 4 main courses =

We have to choose 1 dessert out of 3 desserts available =

So total ways =

Describe your changes (optional) 200